What is the thickness of the atmosphere?

  • #1
144
1

Main Question or Discussion Point

I was thinking of a Fermi-question: the thickness of atmosphere with diffusive equilibrium. And I estimated roughly 10^{5}m (where it should be ~10^{4}m). The difference of order of magnitude to real thickness is 1 (from Wiki).

I had a lot of fun, and I am looking for interesting ways other than chemical potential to estimate the thickness of earth's atmosphere.

---

Basically, I started with assumptions:

- thermal equilibrium
- diffusive equilibrium

such that I can set up the difference in chemical potential is zero, that is equal to the difference in external chemical potential (gravitational potential) plus the difference in internal potential (assuming air molecular just ideal gas); mathematically:

$$\Delta\mu=0=\Delta\mu_{ext}+\Delta\mu_{int}$$

which is

$$0=\Delta\mu_{gra}+\Delta\mu_{indeal}$$

$$0=-GM_{earth}m\Big( \frac{1}{R+\Delta h} - \frac{1}{R} \Big)+k_BT\ln\Big(\frac{n_2}{n_1}\Big)$$

Setting the ratio of n_2, n_1 to be 0.1 (so I defined the height where density of air goes from unity - ground - to 0.1 as "thickness"), and m the mass of air molecular to be 10^{-26}kg, T to be 300K, then we will eventually obtain: $$\Delta h \approx 10^5m$$
 
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Answers and Replies

  • #2
34,477
10,600
You underestimate the mass of a molecule in the air by a factor 6, that leads to most of the difference.
There are many ways to get the same expression, but they all lead to good estimates.
 
  • #3
144
1
You underestimate the mass of a molecule in the air by a factor 6
Thanks for telling me!

There are many ways to get the same expression, but they all lead to good estimates.
I am all ears :)
 
  • #4
34,477
10,600
You can calculate the kinetic energy of a gas molecule, and see how high it can get if it moves straight up with this energy.
You can calculate the ratio of pressure difference to pressure needed to balance a pocket of gas against gravity.
And probably a few more.

They all boil down to ##h=\frac{kT}{mg} \approx 9\,km## with additional prefactors of ln(2) or similar depending on what you do.
 
  • #5
300
11
I would tend to think that the atmosphere ends in the same way the heliosphere ends - i.e., where the particles in intersolar space are in equilibrium with the particles escaping Earth. Of course, Von Karman had a great definition in which a standard aircraft body would need to be going so fast to maintain altitude that it would be in orbit.
 

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